I am unsure how to proceed with the current equation to determine a three-term outer expansion and three-term inner expansion due to the nature of the equation.
Equation:
$\epsilon \frac{d^2y}{dx^2}+\frac{dy}{dx} = 2x$, $y(0) = 0$, $y(1) = 2$
Whilst I only have lecture notes with equations in the form:
$\epsilon \frac{d^2y}{dx^2}+ a(x)\frac{dy}{dx} + b(x)y = 0$
I have made an attempt to a solution:
Seeking a solution in the form of
$y(x) = y_0(x) + \epsilon y_1(x) + \epsilon^2 y_2(x) + O(\epsilon^3)$
$=> \epsilon (y_0^{''} + \epsilon y_1^{''} + \epsilon^2 y_2^{''} + O(\epsilon^3)) + y_0^{'} + \epsilon y_1^{'} + \epsilon^2 y_2^{'} + O(\epsilon^3) = 2x$
Equating powers of $\epsilon$
$\epsilon^0: y_0 = 2x^2$
$\epsilon^1: y_1 = 6x^2 - 4x$
$\epsilon^2: y_2 = 14x^2 - 12x$
$y_{outer}(x) = 2x^2 + \epsilon (6x^2 - 4x) + \epsilon^2 (14x^2 - 12x)$
applying the transform $y(x) = Y(X)$ and $x = \epsilon X$ we get:
$\frac{d^2y}{dx^2}+\frac{dy}{dx} = 2x\epsilon$
again seeking a solution in the form of
$Y_{inner}(X) = Y_0 + \epsilon Y_1 + \epsilon^2 Y_2$
however I am stuck here when attempting to use the following formula:
$\frac{d^2Y_n}{dX} + \frac{dY_n}{dX} = 2x - \frac{dY_{n-1}}{dX} - Y_{n-1}$
Thanks in advance